# Video: AQA GCSE Mathematics Foundation Tier Pack 3 • Paper 3 • Question 9

Perry has some regular shapes. Quentin also has some regular shapes. Perry gives three of her shapes to Quentin. Now, the total number of sides of Perry’s and Quentin’s shapes is the same. Which three shapes did Perry give to Quentin?

04:12

### Video Transcript

Perry has some regular shapes. Quentin also has some regular shapes. Perry gives three of her shapes to Quentin. Now, the total number of sides of Perry’s and Quentin’s shapes is the same. Which three shapes did Perry give to Quentin?

To find the number of shapes that Perry gave to Quentin, we’ll need to consider the number of sides that each of these shapes have. So we’ll label our triangles with a three, our square four, the pentagon five, and the hexagons six. The same thing goes for Quentin shapes. At this point, we’ll total the number of sides that Perry started with and the number of sides that Quentin started with. Perry’s triangles total to nine sides. When we combine the square and the pentagon, we get nine sides. And then, if we multiply six times three, we get 18. Nine plus nine plus 18 equals 36. Perry started with 36 sides.

If we add Quentin shapes, three plus five equals eight plus six equals 14. Quentin started with 14 sides. After Perry gave away three shapes to Quentin, they had the same amount of sides. We can calculate this amount by finding the midpoint between 14 and 36. To do that, we’ll add the sides together, 36 plus 14 and then divide that value by two. 36 plus 14 is 50. So we divide 50 by two and we get 25. This means Perry is going from having 36 sides to having 25 sides and Quentin moved from 14 sides to 25 sides. Perry lost 11 sides and Quentin gained 11 sides. That means we’re looking for three shapes with a total of 11 sides.

We now need to consider which combination of Perry’s shapes will total to 11. If we look at the three triangles, they add together to equal nine sides and that’s not enough. The three hexagons together add up to 18 sides and that’s too many. If we look at the square and the pentagon together, they have nine. If we tried to add one triangle to this, we will get nine plus three which equals 12. If three plus four plus five equals 12, that’s one too many sides than what we’re looking for. Since three, four, and five were one too many, we can replace the square that has four sides with the triangle that has three. Three plus three plus five equals 11.

If Perry gave Quentin two triangles and a pentagon, he loses 11 sides and Quentin gains 11 sides. After which, they would both end up with 25 sides. For the solution, we’ll list out the three shapes as triangle, triangle, pentagon. Our first strategy for solving this problem involved the midpoint.

I want to show one other way we can solve this problem. We could have calculated the difference between Perry’s sides and Quentin’s sides. 36 minus 14 equals 22. And then divide that difference by two. Take half the difference. 22 divided by two equals 11, which would have told us that Perry need to give away 11 sides.

Either by finding the difference and then having that value or by finding the midpoint, you can find that Perry needs to give Quentin two triangles and a pentagon.